1) Find the area of the region bounded by the graphs of
f(x) = x^2 - 5x + 3
and
H(x) = -x

Is H(x) an antiderivative? how do I solve this?

It looks like you're meant to assume that H(x) is just another function, like the quadratic you're given. What you need to do is find out where H(x) = f(x) and then find the areas under H(x) and f(x) between these limits and subtract one from the other. Finding the area under H(x) won't even require integration, as it's a straight line.

Did you draw a diagram so that you could see what was going on? You could simply plug the two functions into your computer or graphing calculator and see what bounded region the question is talking about. But it's not hard to graph it by hand, right?

h(x) = -x --a line passing through the origin with slope -1

The second graph is obviously parabolic, and it's easy to see what it's like by completing the square:

[tex] f(x) = x^2 - 5x + 3 = x^2 - 5x + 25/4 + 3 - 25/4 [/tex]

[tex] = (x - 5/2)^2 -13/4 [/tex]

So the graph is just that of y = x2 with a vertex shifted 5/2 to the right and 3 1/4 down. Now that you're looking at the graph, you can see that it's bound by h(x) on top and f(x) below. Can you see then that the area will be given by:

[tex] \int_a^b {(h(x) - f(x))}dx [/tex]

But what are these points bounding the region on the left and right, with x-coordinates a and b, respectively? They are the intersection points of the graphs of the two functions! They must satisfy both expressions, so that:

[tex] -x = x^2 - 5x + 3 [/tex]

[tex] 0 = x^2 - 4x + 3 = (x-3)(x-1) [/tex]

[tex] x = 1, x = 3 [/tex]

so we have intersection points (1, -1) and (3, -3)

The area becomes:

[tex] \int_1^3 {(-x - x^2 + 5x -3)}dx [/tex]

[tex] = \int_1^3 {(- x^2 + 4x -3)}dx [/tex]

I get -4/3 by hand, which agrees with the TI-83 answer -1.33333333333333....